Question

Find the remainder when x^23 is divided by x²-3x+2​

Answer 1

GIVEN

P(x) = x^23

g(x) = x^2 - 3x + 2

TO FIND :- Remainder when p(x) is divided by g(x)

SOLUTION

The degree of p(x) is 23 which is more fir performing long division method. Therefore we will go for Remainder Theorem.

According to the remiander theorem when we will substitute the zero of g(x) in place of x in p(x) we will get the remainder.

Zero of g(x) by middle term factorisation

${x}^{2} - 3x + 2 = 0 \\ \\ \\ \\ \implies {x}^{2} - 2x - x + 2 = 0 \\ \\ \\ \\ \implies x(x - 2) - 1(x - 2) = 0 \\ \\ \\ \\ \implies (x - 2)(x - 1) = 0 \\ \\ \\ \\ \implies \boxed{x = 2 \: or \: x = 1}$

Substituting value of x in p(x)

p(x) = x^23

=> p(1) = 1^23 = 23

OR

p(x) = x^23

=> p(2) = 2^23 = 8,388,608

We can consider 1 as remainder when p(x) is divisible by g(x) . [ANS]

Answer 2

Step-by-step explanation:

Given: x^23 = Dividend

            x^2-3x+2 = divisor

           Assuming quotient= q(x)

            Remainder = ax+b  (taken)

ATC, x^23= (x^2-3x+2)*q(x)+ax+b

or, x^23= (x-1)(x-2)*q(x)+ax+b

             x=1,2 (from equating the dividend)

Taking x=1 from the value of x found earlier,

1^23=(1-1)(1-2)*1+ax+b

or, 1= a+b..................(case i)

Taking x=2,

2^23= (2-1)(2-2)*2+2a+b

or, 2^23= 2a+b............(case ii)

a=1-b from case (i)

2a=2^23-b from case (ii)

subtracting case (ii) from case(i)

a=2^23-1 case (iii)

taking case (i),

a=1-b

b=1-a

b=1-(2^23-1)

b=2-2^23 case (iv)

now value of remainder(ax+b)=(2^23-1)x+2-2^23 (using case iii and iv)

=(2^23-1)x+(2-2^23) <ans>

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